Combustion enthalpy sign and scaling:\nFor the reaction C6H6 + 7.5 O2 → 6 CO2 + 3 H2O with ΔH = 3264.6 kJ per kg·mole of benzene, what is the heat change when 39 g of benzene burns in an open container?

Introduction / Context:
Heat (enthalpy) of reaction values must be applied with correct stoichiometric scaling and sign convention. Combustion reactions are exothermic, so the enthalpy change is negative for the system at constant pressure (open container). This question checks proportional scaling from a tabulated per–kg·mole value to the amount actually burned, and the correct sign.

Given Data / Assumptions:

• ΔH given as 3264.6 kJ per kg·mole of benzene (interpreted as magnitude).
• Molar mass of benzene = 78 g/mole.
• Mass burned = 39 g = 0.5 mole = 0.0005 kg·mole.
• Open container ⇒ at ~constant pressure, heat released equals enthalpy change.

Concept / Approach:
Scale the tabulated enthalpy by the extent of reaction. Because combustion is exothermic, the sign is negative for the system. Here, 39 g corresponds to half a mole, i.e., 0.5 of the per-mole (or per–kg·mole) basis, so the magnitude halves. Ensure units are understood consistently with the provided data format; the numeric scaling is unaffected by the “kg·mole” label when comparing half a mole to one mole.

Step-by-Step Solution:

Verification / Alternative check:
Using the concept that ΔH scales linearly with reaction extent, any partial burning yields the corresponding fraction of the tabulated heat, preserving the exothermic sign.

Why Other Options Are Wrong:

• Positive values imply endothermic behavior, which is incorrect for combustion.
• -2448.45 kJ corresponds to 0.75 of the tabulated value, not 0.5.
• Zero heat is unphysical for combustion.

Common Pitfalls:
Dropping the sign of ΔH, mixing per-mole vs per–kg·mole bases, or forgetting to scale for the actual number of moles burned.

Final Answer:
-1632.3 kJ/kg·mole.